我做了个保健品站 http://www.bjp51.net 这两天有点问题，用360一扫，高危，只有49分，报告如下：
第一个是很严重的问题，今天想把第一个问题解决了。
看下360给出的意见
我不知道怎么改，希望大家帮忙解决下。
回复讨论(解决方案)
这是contrast.php的文件源码
<body><?php$id='';if(!empty($_POST['id'])){ for($i=0; $i<count($_POST['id']);$i++){ $id=$id.($_POST['id'][$i].','); } $id=substr($id,0,strlen($id)-1);//去除最后面的","}if ($id==''){echo "<script lanage='javascript'>alert('操作失败！至少要选中一条信息。');window.opener=null;window.open('','_self');window.close()</script>";exit;} $tdwidth=floor(90/$i);//取整,左边占10%$sql="select * from zzcms_main where id in ($id)" ;$rs=mysql_query($sql);?><table width="100%" height="218" border="0" align="center" cellpadding="5" cellspacing="1" class="bgcolor3"> <tr> <td width="10%" align="center" bgcolor="#FFFFFF">【产品图片】 </td> <?php while ($row=mysql_fetch_array($rs)){?> <td bgcolor="#FFFFFF" style="font-weight:bold" width="<?php echo $tdwidth ?>%"><a href="<?php echo $row["img"]?>" target="_blank"><img <?php echo getsmallimg($row["img"],"")? alt="网站出问题了，有漏洞，今天来这里请csdn的兄弟帮忙解决" > alt="<?php echo $row["proname"]?>" border="0" ></a></td> <?php } ?> </tr> <tr class="bgcolor1"> <td width="100" align="center">【产品名称】 </td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td style="max-width:90%"><?php echo $row["proname"]?></td> <?php } ?> </tr> <tr> <td width="100" align="center" bgcolor="#FFFFFF">【主要功能】</td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td valign="top" bgcolor="#FFFFFF" ><?php echo $row["prouse"]?></td> <?php } ?> </tr> <tr class="bgcolor1"> <td width="100" align="center">【规格包装】</td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td><?php echo $row["gg"]?></td> <?php } ?> </tr> <tr class="bgcolor1"> <td width="100" align="center"><strong>招商区域</strong></td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td><?php echo $row["city"]?></td> <?php } ?> </tr> <tr> <td width="100" align="center" bgcolor="#FFFFFF"><strong>产品说明</strong></td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td valign="top" bgcolor="#FFFFFF"><?php echo nl2br($row["sm"])?></td> <?php } ?> </tr> <tr class="bgcolor1"> <td width="100" align="center"><strong>可提供的支持</strong></td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td valign="top"><?php echo nl2br($row["zc"])?></td> <?php } ?> </tr> <tr> <td width="100" align="center" bgcolor="#FFFFFF"><strong>对代理商的要求</strong></td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td valign="top" bgcolor="#FFFFFF"><?php echo nl2br($row["yq"])?></td> <?php } ?> </tr> <tr class="bgcolor1"> <td width="100" align="center"><strong>备注</strong></td> <?php mysql_data_seek($rs,0); while ($row=mysql_fetch_array($rs)){?> <td><?php echo nl2br($row["other"])?></td> <?php } ?> </tr></table><table width="100%" style="max-width:90%" border="0" cellpadding="0" cellspacing="0"> <tr> <td align="center"><input type="button" name="Submit" value="关闭本窗口" onClick="javascript:window.close()"></td> </tr></table><?phpmysql_close($conn);?></body> $id='';
if(!empty($_POST['id'])){
for($i=0; $i
}
$id=substr($id,0,strlen($id)-1);//去除最后面的","
}
$sql="select * from zzcms_main where id in ( $id)"
他认为你未经检查就在 sql 指令中使用了传入的数据
$id='';
if(!empty($_POST['id'])){
for($i=0; $i
}
$id=substr($id,0,strlen($id)-1);//去除最后面的","
}
$sql="select * from zzcms_main where id in ( $id)"
他认为你未经检查就在 sql 指令中使用了传入的数据
我是php小白，请问我应该怎么验证，怎么检查好呢？
还有我不检查，是不是很容易被注入呢？
$id = $id . (intval($_POST['id'][$i]) . ',');
SQL注入的原理是，从地址栏或者表单中注入
如果你从地址栏得到一个$_GET["a"]，不经过过滤就直接使用到程序中，就会造成威胁。比如：
如果$_GET["a"]=1；那么:
$sql = "SELECT * FROM AA WHERE id =$_GET["a"]";就是$sql = "SELECT * FROM AA WHERE id =1";
但如果别人通过地址栏自行修改，把$_GET["a"]的值改为1 or （and） XXX各类代码，那这个查询语句就变成
$sql = "SELECT * FROM AA WHERE id =1 or（and） xxx";
于是就中招了。
所以地址栏和表单得到的参数，一定要格式化，过滤好，指定是什么类型，多长，限制哪些字符……
$sql="select * from zzcms_main where id in ($id)" ;
$id?有?行??，用??入什?都可以，?然被注入了。
因?id只能是?字，所以可以用intval?成?字，如果非?字???0，??就注入不到了。
安全级别：高危
安全等级打败了全国46%的网站！
$id=$id.(intval($_POST['id'][$i]).',');
传入的数据把单引号替换为两个连续的单引号 ， sql语句用传入的参数时加上单引号。
$id = str_replace("'","''",$_POST['id']);
$sql = " select * from tb_user wher id='$id' ";
这样就不怕注入了。
谢谢 您们的回答 我目前是把$id=$id.($_POST['id'][$i].','); 替换成 $id = $id . (intval($_POST['id'][$i]) . ','
就对了
上面的不对，我改成这样了
if(!empty($_POST['id'])){
for($i=0; $i
$id = $id.(intval($_POST['id'][$i]).',');
}
$id=substr($id,0,strlen($id)-1);//去除最后面的","
}
这样是不是对了
if(!empty($_POST['id'])) { $id = join(',', array_map('intval', $_POST['id']));}最好把 $id 换个名字
if(!empty($_POST['id'])) { $id = join(',', array_map('intval', $_POST['id']));}最好把 $id 换个名字 我对php不熟，我就用这段代码，不用找变量了
感谢版主无私的精神。
我才发现，另一个问题也是您在为我解决，再次感谢！
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